若极限lim((3n^2+4)/(n+2)-an+b)=6,则a=_,b=_
若极限lim((3n^2+4)/(n+2)-an+b)=6,则a=_,b=_
lim (n→∞) (n^2/(an+b)-n^3/(2n^2-1))=1/4 求a,b
若lim[2n+(an^2+2n+1)/(bn+1)=1,则a+b
若lim[(an^2+bn+c)/(2n-3)]=-2,则a+b=
求数列极限lim=[(an^2+bn-1)/(4n^2-5n+1)]=1/b 求a b的值
几道大一 求极限:1.lim(√n -9)/(n+3)=2.lim(1+a+a^2+a^3+.+a^n)/(1+b+b^
数列极限基本题已知数列{an}的极限为0,且有lim[(3n-2)an]=6,则lim[n(an)]=?
lim (n→∞) [(an^2+bn+c)/(2n+5)]=3,求a,b
已知lim[(an^2+5n-2)/(3n+1)-n]=b,求a,b的值
(n²+1)/(n+1)-an-b的极限为0,则a+2b=
lim(n2+2n+2)/(n+1)-an)=b,求a,b
已知:lim (n→∞) [(n^2+n)/(n+1)-an-b]=1 ,求a,b的值